Why Choose Us?
0% AI Guarantee
Human-written only.
24/7 Support
Anytime, anywhere.
Plagiarism Free
100% Original.
Expert Tutors
Masters & PhDs.
100% Confidential
Your privacy matters.
On-Time Delivery
Never miss a deadline.
A parallel-plate capacitor is charged to 5000 V
A parallel-plate capacitor is charged to 5000 V. A proton is fired into the center of the capacitor at a speed of 3.7×105 m/s as shown in (Figure 1) . The proton is deflected while inside the capacitor, and the plates are long enough that the proton will hit one of them before emerging from the far side of the capacitor.
Expert Solution
E = electric field between the plates
V = potential difference = 5000 v
d = distance between the plates
Electric field is given as
E = V/d = 5000/d
q = charge on proton
Force on the proton
F = qE
a = acceleration = F/m = qE/m = 5000q/(md)
Y = displacement = d/2
initial velocity = Vi = 0
Vf = final velocity
Using the equation
Vf2 = Vi2 + 2 a Y
Vf2 = 02 + 2 (5000q/(md)) (d/2)
Vf2 = 5000 x 1.6 x 10-19 / (1.67 x 10-27)
Vf2 = 4.8 x 1011
Along the horizontal direction
Vf' = Vi = 3.7 x 105 m/s
impact speed = V = sqrt (Vf2 + Vf'2 ) = sqrt (4.8 x 1011 + (3.7 x 105 )2) = 7.86 x 105 m/s
Archived Solution
You have full access to this solution. To save a copy with all formatting and attachments, use the button below.
For ready-to-submit work, please order a fresh solution below.
